B=1/2.1.2-1/2.2.3+1/2.2.3-1/2.3.4+...+1/2n(n+1)-1/2(n+1)(n+2)

B=1/2[(1/1.2+1/2.3+...+1/n(n+1))-(1/2.3+1/3.4+...+1/(n+1)(n+2))]

Tới đây bạn tự làm tiếp nha, tương tự như bài 1/1.2+1/2.3+..+1/n(n+1) á bạn.Cái này bạn ghi ra bạn sẽ hiểu, mình viết hơi bị lủng củng.

99.101 mới đúg nhé

=\(\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{10000}{99.101}\)

=\(\frac{2^2.3^2.4^2......100^2}{\left(1.2.3.....99\right).\left(3.4.5.....101\right)}=\frac{\left(2.3.4....100\right).\left(2.3.4....100\right)}{\left(1.2.3....99\right).\left(3.4.5......101\right)}\)

=\(\frac{100.2}{1.101}=\frac{200}{101}\)

A=\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)..........\left(\frac{2017.2019+1}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.............\frac{4072324}{2017.2019}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...................\frac{2018^2}{2017.2019}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{\left(2.3.4..........2018\right).\left(2.3.4............2018\right)}{\left(1.2.3............2017\right).\left(3.4.5..........2019\right)}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2018.2}{1.2019}\right)=\frac{2018.2}{2.2019}=\frac{2018}{2019}\)

Vậy \(A=\frac{2018}{2019}\)

Chúc bn học tốt

\(A:\frac{1}{2}=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2017.2019+1}{2017.2019}\)

\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{2018^2}{2017.2019}\)

\(=\frac{2.2.3.3.4.4.....2018.2018}{1.3.2.4.3.5....2017.2019}\)

\(=\frac{2.3.4.....2018}{1.2.3.4.....2017}.\frac{2.3.4....2018}{3.4.5.....2019}\)

\(=2018.\frac{2}{2019}\)

\(=\frac{4036}{2019}\)

\(\Rightarrow A=\frac{4036}{2019}.\frac{1}{2}\)

\(A=\frac{2018}{2019}\)